tìm x biết
(1/1.2.3+1/2.3.4+1/3.4.5+...+1/8.9.10)x=23/45
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
thay A vào ta được
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
Đặt A=11.2.3+12.3.4+....+18.9.10A=11.2.3+12.3.4+....+18.9.10
2A=21.2.3+22.3.4+....+28.9.102A=21.2.3+22.3.4+....+28.9.10
=3−11.2.3+4−22.3.4+...+10−88.9.10=3−11.2.3+4−22.3.4+...+10−88.9.10
=11.2−12.3+12.3−13.4+...+18.9−19.10=11.2−12.3+12.3−13.4+...+18.9−19.10
=11.2−19.10=2245=11.2−19.10=2245
A=1145A=1145
Ax=1145x=2245Ax=1145x=2245
x=2245:1145=2
Ta có: \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
Đặt A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}\) \(+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
2A=\(\frac{1}{1.2}-\frac{1}{9.10}\)
2A=\(\frac{22}{45}\)
A=\(\frac{22}{45}\div2\)
A=\(\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy x=\(\frac{23}{11}\)